Find the distance between the points :

Solution:

Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by

$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

(i) $A(5,1,2)$ and $B(4,6,-1)$

Here, $\left(x_{1}, y_{1}, z_{1}\right)=(5,1,2)$

$\left(x_{2}, y_{2}, z_{2}\right)=(4,6,-1)$

Therefore,

$D=\sqrt{(4-5)^{2}+(6-1)^{2}+(-1-2)^{2}}$

$=\sqrt{(-1)^{2}+(5)^{2}+(-3)^{2}}$

$=\sqrt{1+25+9}$

$=\sqrt{35}$

Distance between points A and B is

$\sqrt{35}$

(ii) $P(1,-1,3)$ and $Q(2,3,-5)$

Here, $\left(x_{1}, y_{1}, z_{1}\right)=(1,-1,3)$

$\left(x_{2}, y_{2}, z_{2}\right)=(2,3,-5)$

Therefore,

$D=\sqrt{(2-1)^{2}+(3-(-1))^{2}+(-5-3)^{2}}$

$=\sqrt{(1)^{2}+(4)^{2}+(-8)^{2}}$

$=\sqrt{1+16+64}$

$=\sqrt{81}=9$

Distance between points P and Q are 9 units

(iii) $R(1,-3,4)$ and $S(4,-2,-3)$

Here, $\left(x_{1}, y_{1}, z_{1}\right)=(1,-3,4)$

$\left(x_{2}, y_{2}, z_{2}\right)=(4,-2,-3)$

Therefore,

$D=\sqrt{(4-1)^{2}+(-2-(-3))^{2}+(-3-4)^{2}}$

$=\sqrt{(3)^{2}+(1)^{2}+(-7)^{2}}$

$=\sqrt{9+1+49}$

$=\sqrt{59}$

Distance between points $R$ and $S$ is $\sqrt{59}$ units.

(iv) $C(9,-12,-8)$ and the origin

Coordinates of origin are $(0,0,0)$

Here, $\left(x_{1}, y_{1}, z_{1}\right)=(9,-12,-8)$

$\left(x_{2}, y_{2}, z_{2}\right)=(0,0,0)$

Therefore,

$D=\sqrt{(0-9)^{2}+(0-(-12))^{2}+(0-(-8))^{2}}$

$=\sqrt{(-9)^{2}+(12)^{2}+(8)^{2}}$

$=\sqrt{81+144+64}$

$=\sqrt{289}=17$

Distance between points $C$ and origin is 17 units.